Vì \(abc=105\)
\(S=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)
\(S=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(b+1+bc\right)}\)
\(S=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)
\(S=\frac{bc+b+1}{bc+b+1}=1\)
Vậy \(S=1\)